Show that $x^2 + 2y^2 + z^2 - 2yz > 0$.

Question

Let $(x, y, z) \in \mathbb{R}^3 - \{0\}$.

Show that $x^2 + 2y^2 + z^2 - 2yz > 0$.

I've been pushing symbols around for a while on this one. I've tried hitting it with all of the "named" inequalities. It's probably something extremely simple that I'm overlooking.

Answer 1

$x^2 + 2y^2 + z^2 - 2yz = x^2+y^2+(y-z)^2>0$ for $(x, y, z) \in \mathbb{R}^3 \setminus \{0\}$. Since squares are always non negative.

$\textbf{Note:}$ $x^2+y^2+(y-z)^2\neq0$ for if that were the case then $x^2,y^2,(y-z)^2=0$ i.e. $(x,y,z)=(0,0,0)$, a contradiction to our assumption $:(x, y, z) \in \mathbb{R}^3 \setminus \{0\}$